Does a net converging to a point $x$ have a countable subset $\{x_n\}_n$ such that it converges to $x$ in a sequentially compact space?

Question

I was thinking for example, in functional analysis, we can talk about a Banach space $E$ and it's dual space $E^{*}$ with it's $w^{*}$-topology. In the special case that $E$ is infinite dimensional and reflexive, we can say that the $w^{*}$-topology in $E^{*}$ is the same as the weak topology. I also know that because $E^{*}$ must be infinite dimensional, we have that $S := \{\phi \in E^{*} : \left \| \phi \right \| = 1 \}$ has its topological closure equal to $B_{E^{*}}:=\{ \phi \in E^{*} : \left \| \phi \right \| \leq 1 \}$ in the weak topology. What is more, $B_{E^{*}}$ is compact in the $w*$-topology (which is the same as the weak topology) by Banach-Alaoglu theorem (one can argue the same thing saying that $E^{*}$ is also reflexive), but by Eberlein–Šmulian theorem it is sequentially compact. I would like to know if this implies that, because I have a net of elements in $S$ converging to $0$ weakly, this implyies that there is a sequence of elements in $S$ converging to $0$ weakly.

I have the feeling that sequential compactness doesn't imply that. Probably because this is similar to say that every infinite dimensional reflexive space has Schur's property, which is not true, for example, if we take an infinite dimensional separable hillbert space $H$ (in particular reflexive), because then it's dual is separable and infinite dimensional which implies (exercise) that $H$ doesn't have Shur's property.

Answer 1

A counterexample is the ordinal space $\omega_1+1$. This is sequentially compact, since every sequence either has a constant subsequence or else is bounded below $\omega_1$ and so has a convergent subsequence. But the identity net on $\omega_1$ converges to $\omega_1$, and no countable subset forms a sequence converging to $\omega_1$.

Answer 2

Here are two examples in the Hilbert space $\ell^2$. Take the set $$A = \{\sqrt{n} \, e_n \;\mid\; n \in \mathbb{N}\}.$$ Then, you can show that $0$ belongs to the weak closure of $A$, i.e., there is a net in $A$ converging to $0$. However, $A$ is sequentially closed.

Similarly, you can take $$B = \{ e_n + n \, e_m \;\mid\; n,m \in \mathbb{N}, n < m\}.$$ Again, $0$ is in the weak closure of $B$, but no subsequence converges weakly to $0$.

(I think it is also possible to write down some nets explicitely.)